有沒有更好的方法來根據條件（Google Apps Script）合並行數據？

Question

好的...首先讓我說我是自學的 Google Apps 腳本...說的夠多了，對吧！？ 下面的腳本正在運行，但我想優化它或想出另一種方法來實現相同的結果。 該腳本采用 18000 行和 86 列數據，並根據 id 列表將它們組合成單行。 id 列表大約有 13000 行。 簡短的版本是這樣的……它按 id 過濾數據，然后檢查每一列的最后一行是否有提交的數據並返回該單元格。 例如：

//sample data
[[311112, 1, 2, 4, 5,"","","","","","", 2, 3],
[323223,"","","","","", 2, 4, 4,"","","",""],
[321321, 1, 2, 4, 5,"","","","","","", 2, 3],
[311112, 4, 1, 6, 7,"", 3,"", 3,"","", 5, 3],
[321233,"","","","","","", 4, 3, 1, 5,"",""],
[321321,"","","","","","","","", 1 ,4,"",""],
[323223,"","","","","", 2, 3,"","","","",""],
[323153,"", 2, 3, 6,"","","","","","","",""],
[321321,"","","","","", 2, 3,"","","","",""],
[321321,"", 5, 3,"", 1,"","","","","","",""]]

//filtered Data by id 321321
[[321321, 1, 2, 4, 5,"","","","","","", 2, 3],
[321321,"","","","","","","","", 1, 4,"",""],
[321321,"","","","","", 2, 3,"","","","",""],
[321321,"", 5, 3,"", 1,"","","","","","",""]]

// returned row is getting the last nonempty value for each column from the filtered data.

[[321321, 1, 5, 3, 5, 1, 2, 3,"", 1, 4, 2, 3]]

腳本完成大約需要16-18 分鍾。 有沒有更好的方法來完成這個或任何優化建議？

function combineR(startRow, startRange) {
  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var sheets = ss.getSheets();
  var testSheet = ss.getSheetByName('Raw Scores');
  var cSheet = ss.getSheetByName('Combined Scores');
  var gradingResults = testSheet.getRange(1, 1, testSheet.getLastRow(), testSheet.getLastColumn()).getValues();

  if (startRow > cSheet.getLastRow()) {
    return;
  }

  if (startRow + startRange > cSheet.getLastRow()) {
    startRange = cSheet.getLastRow() - startRow;
  }

  var sID = cSheet.getRange(startRow, 2, startRange).getValues();
  var maxScores = [];
  for (var x = 0; x < sID.length; x++) {
    var filtered = gradingResults.filter(function (dataRow) {
      return dataRow[0] === sID[x][0];
    });

    if (isFinite(filtered)) {
      maxScores.push(['', '', '', '', '', '', '', '', '', '',
        '', '', '', '', '', '', '', '', '', '',
        '', '', '', '', '', '', '', '', '', '',
        '', '', '', '', '', '', '', '', '', '',
        '', '', '', '', '']);
      continue;
    } else {
      maxScores.push(['', getMaxLetter(filtered, 3), lastGraded(filtered, 4), lastGraded(filtered, 5), lastGraded(filtered, 6), lastGraded(filtered, 7), lastGraded(filtered, 8), lastGraded(filtered, 9), lastGraded(filtered, 10), lastGraded(filtered, 11),
        lastGraded(filtered, 12), lastGraded(filtered, 13), lastGraded(filtered, 14), lastGraded(filtered, 15), lastGraded(filtered, 16), lastGraded(filtered, 17), lastGraded(filtered, 18), lastGraded(filtered, 19), lastGraded(filtered, 20), lastGraded(filtered, 21),
        lastGraded(filtered, 22), lastGraded(filtered, 23), lastGraded(filtered, 24), lastGraded(filtered, 25), lastGraded(filtered, 26), lastGraded(filtered, 27), lastGraded(filtered, 28), lastGraded(filtered, 29), lastGraded(filtered, 30), lastGraded(filtered, 31),
        lastGraded(filtered, 32), lastGraded(filtered, 33), lastGraded(filtered, 34), lastGraded(filtered, 35), lastGraded(filtered, 36), lastGraded(filtered, 37), lastGraded(filtered, 38), lastGraded(filtered, 39), lastGraded(filtered, 40), lastGraded(filtered, 41),
        lastGraded(filtered, 42), lastGraded(filtered, 43), lastGraded(filtered, 44), lastGraded(filtered, 45), lastGraded(filtered, 46)]);
    }
  }
  cSheet.getRange(startRow, 11, maxScores.length, maxScores[0].length).setValues(maxScores)
}

function getMaxLetter(arr, idx) {
  var letter = arr.map(function (e) { return e[idx] }).sort().pop();
  return letter;
}

function lastGraded(arr, idx) {
  var newArray = arr.map(function (e) { return e[idx] });
  newArray.reverse();
  for (var x = 0; x < newArray.length; x++) {
    if (typeof newArray[x] == 'number') {
      return newArray[x];
    }
  }
  return '';
}

A列有重復的ID需要合並

B 列具有作為最終合並產品的唯一值

Answer 1

問題：

該腳本似乎有各種問題，但主要問題似乎是使用各種索引多次調用lastGraded函數。 這會為每個索引map 、 reverse和其他所有內容，並且會花費時間。

解決方案：

鑒於您的樣本數據，我建議采用以下方法：

• 獲取 1 個二維數組中的所有輸入數據

• 將輸入數據減少到Map 。 地圖將每個id作為key並將與該key匹配的所有行作為每個key二維array 。 這將以內存為代價大大提高性能/速度。 這比按每個 id 過濾數組要好，因為，

  • 您只循環輸入數組一次
  • 而arr.filter必須為每個 id 循環數組

• 一旦縮減為一個map ，對最后一行中的每個元素反向循環遍歷map中的每個array以找到非空元素。

示例片段：

 const arrMain = //sample data [ [311112, 1, 2, 4, 5, '', '', '', '', '', '', 2, 3], [323223, '', '', '', '', '', 2, 4, 4, '', '', '', ''], [321321, 1, 2, 4, 5, '', '', '', '', '', '', 2, 3], [311112, 4, 1, 6, 7, '', 3, '', 3, '', '', 5, 3], [321233, '', '', '', '', '', '', 4, 3, 1, 5, '', ''], [321321, '', '', '', '', '', '', '', '', 1, 4, '', ''], [323223, '', '', '', '', '', 2, 3, '', '', '', '', ''], [323153, '', 2, 3, 6, '', '', '', '', '', '', '', ''], [321321, '', '', '', '', '', 2, 3, '', '', '', '', ''], [321321, '', 5, 3, '', 1, '', '', '', '', '', '', ''], ]; //reduce input array to a map of id=>rows const map = arrMain.reduce((map, row) => { if (!map.has(row[0])) map.set(row[0], [row]); else map.get(row[0]).push(row); return map; }, new Map()); const out = []; map.forEach(arr2d => { const l = arr2d.length - 1, lastRow = arr2d[l].slice(0); //iterate lastrow of this id's column elements for (let j = 0; j < lastRow.length; ++j) { if (lastRow[j] === '') { //iterate each row of this id for (let i = l; i >= 0; --i) { if (arr2d[i][j] !== '') { lastRow[j] = arr2d[i][j]; break; } } } } out.push(lastRow); }); console.log(out);